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Difference between revisions of "2001 AIME II Problems/Problem 15"

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Set the coordinate system so that vertex <math>E</math>, where the drilling starts, is at <math>(8,8,8)</math>. Using a little visualization (involving some [[similar triangles]], because we have parallel lines) shows that the tunnel meets the bottom face (the xy plane one) in the line segments joining <math>(1,0,0)</math> to <math>(2,2,0)</math>, and <math>(0,1,0)</math> to <math>(2,2,0)</math>, and similarly for the other three faces meeting at the origin (by symmetry). So one face of the tunnel is the polygon with vertices (in that order), <math>S(1,0,0), T(2,0,2), U(8,6,8), V(8,8,6), W(2,2,0)</math>, and the other two faces of the tunnel are congruent to this shape.  
 
Set the coordinate system so that vertex <math>E</math>, where the drilling starts, is at <math>(8,8,8)</math>. Using a little visualization (involving some [[similar triangles]], because we have parallel lines) shows that the tunnel meets the bottom face (the xy plane one) in the line segments joining <math>(1,0,0)</math> to <math>(2,2,0)</math>, and <math>(0,1,0)</math> to <math>(2,2,0)</math>, and similarly for the other three faces meeting at the origin (by symmetry). So one face of the tunnel is the polygon with vertices (in that order), <math>S(1,0,0), T(2,0,2), U(8,6,8), V(8,8,6), W(2,2,0)</math>, and the other two faces of the tunnel are congruent to this shape.  
  
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Observe that this shape is made up of two congruent [[trapezoid]]s each with height <math>\sqrt {2}</math> and bases <math>7\sqrt {3}</math> and <math>6\sqrt {3}</math>. Together they make up an area of <math>\sqrt {2}(7\sqrt {3} + 6\sqrt {3}) = 13\sqrt {6}</math>. The total area of the tunnel is then <math>3\cdot13\sqrt {6} = 39\sqrt {6}</math>. Around the corner <math>E</math> and the corner opposite <math>E</math> we're missing an area of <math>6</math>. So the outside area is <math>6\cdot 64 - 2\cdot 6 = 372</math>. Thus the the total surface area is <math>372 + 39\sqrt {6}</math>, and the answer is <math>372 + 39 + 6 = \boxed{417}</math>.
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Observe that this shape is made up of two congruent [[trapezoid]]s each with height <math>\sqrt {2}</math> and bases <math>7\sqrt {3}</math> and <math>6\sqrt {3}</math>. Together they make up an area of <math>\sqrt {2}(7\sqrt {3} + 6\sqrt {3}) = 13\sqrt {6}</math>. The total area of the tunnel is then <math>3\cdot13\sqrt {6} = 39\sqrt {6}</math>. Around the corner <math>E</math> we're missing an area of <math>6</math>, the same goes for the corner opposite <math>E</math> . So the outside area is <math>6\cdot 64 - 2\cdot 6 = 372</math>. Thus the the total surface area is <math>372 + 39\sqrt {6}</math>, and the answer is <math>372 + 39 + 6 = \boxed{417}</math>.
  
 
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Revision as of 17:38, 7 February 2016

Problem

Let $EFGH$, $EFDC$, and $EHBC$ be three adjacent square faces of a cube, for which $EC = 8$, and let $A$ be the eighth vertex of the cube. Let $I$, $J$, and $K$, be the points on $\overline{EF}$, $\overline{EH}$, and $\overline{EC}$, respectively, so that $EI = EJ = EK = 2$. A solid $S$ is obtained by drilling a tunnel through the cube. The sides of the tunnel are planes parallel to $\overline{AE}$, and containing the edges, $\overline{IJ}$, $\overline{JK}$, and $\overline{KI}$. The surface area of $S$, including the walls of the tunnel, is $m + n\sqrt {p}$, where $m$, $n$, and $p$ are positive integers and $p$ is not divisible by the square of any prime. Find $m + n + p$.

Solution

[asy] import three; currentprojection = perspective(5,-40,12); defaultpen(linewidth(0.7)); pen l = linewidth(0.5) + linetype("10 2"); triple S=(1,0,0), T=(2,0,2), U=(8,6,8), V=(8,8,6), W=(2,2,0), X=(6,8,8); draw((1,0,0)--(8,0,0)--(8,0,8)--(0,0,8)--(0,0,1)); draw((1,0,0)--(8,0,0)--(8,8,0)--(0,8,0)--(0,1,0),l); draw((0,8,0)--(0,8,8)); draw((0,8,8)--(0,0,8)--(0,0,1)); draw((8,8,0)--(8,8,6),l); draw((8,0,8)--(8,6,8)); draw((0,8,8)--(6,8,8)); draw(S--T--U--V--W--cycle); draw((0,0,1)--T--U--X--(0,2,2)--cycle); draw((0,1,0)--W--V--X--(0,2,2)--cycle); [/asy]             [asy] import three; currentprojection = perspective(5,40,12); defaultpen(linewidth(0.7)); pen l = linewidth(0.5) + linetype("10 2"); triple S=(1,0,0), T=(2,0,2), U=(8,6,8), V=(8,8,6), W=(2,2,0), X=(6,8,8); draw((1,0,0)--(8,0,0)--(8,0,8),l); draw((8,0,8)--(0,0,8)); draw((0,0,8)--(0,0,1),l); draw((8,0,0)--(8,8,0)); draw((8,8,0)--(0,8,0)); draw((0,8,0)--(0,1,0),l); draw((0,8,0)--(0,8,8)); draw((0,0,8)--(0,0,1),l); draw((8,8,0)--(8,8,6)); draw((8,0,8)--(8,6,8)); draw((0,0,8)--(0,8,8)--(6,8,8)); draw(S--T--U--V--W--cycle); draw((0,0,1)--T--U--X--(0,2,2)--cycle); draw((0,1,0)--W--V--X--(0,2,2)--cycle); [/asy]

Set the coordinate system so that vertex $E$, where the drilling starts, is at $(8,8,8)$. Using a little visualization (involving some similar triangles, because we have parallel lines) shows that the tunnel meets the bottom face (the xy plane one) in the line segments joining $(1,0,0)$ to $(2,2,0)$, and $(0,1,0)$ to $(2,2,0)$, and similarly for the other three faces meeting at the origin (by symmetry). So one face of the tunnel is the polygon with vertices (in that order), $S(1,0,0), T(2,0,2), U(8,6,8), V(8,8,6), W(2,2,0)$, and the other two faces of the tunnel are congruent to this shape.

Observe that this shape is made up of two congruent trapezoids each with height $\sqrt {2}$ and bases $7\sqrt {3}$ and $6\sqrt {3}$. Together they make up an area of $\sqrt {2}(7\sqrt {3} + 6\sqrt {3}) = 13\sqrt {6}$. The total area of the tunnel is then $3\cdot13\sqrt {6} = 39\sqrt {6}$. Around the corner $E$ we're missing an area of $6$, the same goes for the corner opposite $E$ . So the outside area is $6\cdot 64 - 2\cdot 6 = 372$. Thus the the total surface area is $372 + 39\sqrt {6}$, and the answer is $372 + 39 + 6 = \boxed{417}$.

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See also

2001 AIME II (Problems • Answer Key • Resources)
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Problem 14 		Followed by
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